LT3083MPDF datasheet(18/28 Pages) LINER

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LT3083

3083f

–

LT3083

VCONTROL

OUT

VOUT

VIN

3083 F08

SET

RSET

Figure 8. Reducing Power Dissipation Using a Parallel Resistor

The second technique for reducing power dissipation,

shown in Figure 8, uses a resistor in parallel with the

LT3083. This resistor provides a parallel path for current

ﬂow, reducing the current ﬂowing through the LT3083.

This technique works well if input voltage is reasonably

constant and output load current changes are small. This

technique also increases the maximum available output

current at the expense of minimum load requirements.

As an example, assume: VIN = VCONTROL = 5V, VIN(MAX) =

5.5V, VOUT = 3.3V, VOUT(MIN) = 3.2V, IOUT(MAX) = 2A and

IOUT(MIN) = 0.7A. Also, assuming that RP carries no more

than 90% of IOUT(MIN) = 630mA.

Calculating RP yields:

RP =

5.5V − 3.2V

0.63A

= 3.65Ω

(5% Standard Value = 3.6Ω)

The maximum total power dissipation is (5.5V – 3.2V) •

2A = 4.6W. However, the LT3083 supplies only:

2A −

5.5V − 3.2V

3.6Ω

= 1.36A

Therefore, the LT3083’s power dissipation is only:

PDISS = (5.5V – 3.2V) • 1.36A = 3.13W

RP dissipates 1.47W of power. As with the ﬁrst technique,

choose appropriate wattage resistors to handle and dis-

sipate the power properly. With this conﬁguration, the

LT3083 supplies only 1.36A. Therefore, load current can

increase by 1.64A to a total output current of 3.64A while

keeping the LT3083 in its normal operating range.

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